A car travelling at a speed of `30 ms^(-1)` due north along the highway makes a left turn on to a sied road which heads towards due west. It takes `40 s` for the car to complere the turn. At the end of `40 s`, the caar has a speed of `20 ms^(-1)` along the side road. Derermime the magnitude of average acceleration over the `40 second` interval.

Explanation

Refer to fig. 2 (c ) 52, initial velocity ,
` vec v_1 = vec (OA) = 30 ms^(-1)`, due noth. Final velocity
` vec V_2 = vec (OB) =20 ms^(-1)` due west.
.
Change in velocity
` = vec v_2 -vec v-1 = vec (OB) -vec (OA)= vec (AB)`
:. ` | vec v_2- vec -1 | = AB= AB= sqrt ((OA)^2) = sqrt 9(OA)^2 + (OB) ^2 )
` = sqrt 930^2 + 20^2 ) = sqrt ( 1300) = 36 ms^(-1)`
Average accleration,
` a-(av) =|vec v_2 -vec _1| /t = 936)/(40) =0.9 ms^(-2)`.